# Proving a statement regarding a Diophantine equation

FINAL EDIT : Prove that if $p^z|n^2-1$

$$p^{x-z}(p^{z}-1)=\dfrac{ n^2-1}{p^z}-3$$ doesn't hold for any chosen values of $p,x,n$ and $z$.

Here $p>3$ is an odd prime , $x=2y+z, \ \{\{x,y,z\}>0\} \in \mathbb{Z}$ . There $n$ is an even number.

If the above statement is prove it will lead to a contradiction$^*$

$^*$: to understand the contradiction you need to read this :

EDIT : [History] : If anybody remembered the previous question of mine, I asked to prove $(p^x+3)(p^z-1)+4$ is not a perfect square. So I tried this $$p^{x+z}-p^x+3p^z+1=l^2$$ $$p^{x+z}-p^x+3p^{z}=l^2-1$$ $$p^z(p^{x}-p^{x-z}+3)=(l+1)(l-1)$$

Here its evident that $p^z$ divides either of $(l-1)$ or $(l+1)$. So let us assume as case (i) the $p^z|(l-1)$.

So let $k=\large \frac{l-1}{p^z}$, so when $k$ is decimal ( clearly $l-1$ is odd and $p^z$ is odd, so all the time the $k$ is not a integer ) ignore the case as it leads to a contradiction and proves it. Now look when $k$ is integer so the equation $$p^z(p^{x}-p^{x-z}+3)=(l+1)(l-1)$$ can be written as $$(p^{x}-p^{x-z}+3)=k(l+1)$$ $$p^{x}-p^{x-z}=k(l+1)-3.$$ After working on many examples, I have found an interesting pattern between the differences between the same odd number raised to different powers. For suppose we take an odd number $5$ and then work on the difference of the powers of it. So the difference seems to be of the form $O^n-O^{m}(n>m)$ ( where $O$ is an odd number ) . So let us call the set of all such differences $\mathfrak{D}^{n}_{O}$ set of all $\left\{O^n-O^{m}\right\}$ such that the integer $m$ runs from $0$ to $n-1$ . Here For example we can start writing all such differences to see an interesting property.

Fix $O=5$. Let us take and $n=1$ and $\mathfrak{D}^{1}_{5}$ is nothing but the set of 1 element $\left\{5^1-5^0=4\right\}$ Let us now take $n=2$ and $\mathfrak{D}^{2}_{5}$ is nothing but only 2 elements $\left\{5^2-5=20,5^2-5^0=24\right\}$. Let us now take $n=3$. So the $\mathfrak{D}^{3}_{5}$ is nothing but set of 3 elements $\left\{5^3-5^2=100,5^3-5=120,5^3-5^0=124\right\}$ \( Since for $n=3$ there are only two possible $m=1,2 (3-1=2)$. Let us now take $n=4$. So the $\mathfrak{D}^{4}_{5}$ is nothing but the set of 3 elements $\left\{5^4-5^3=500,5^4-5^2=600,5^4-5=620,5^4-5^0=624\right\}$. Let us now take $n=5$. So the $\mathfrak{D}^{5}_{5}$ is nothing but the set of 4 elements $\left\{5^5-5^4=2500,565-5^3=3000,5^5-5^2=3100,5^5-5=3120,5^5-5^0=3124\right\}$.

And so on for different values of $n$. If we observe we find that the elements of the sets follow a good pattern. After trying for many such numbers I came to know the pattern. Let me explain it sir. So let us write down all such $\mathfrak{D}^{n}_{5}$ $$\mathfrak{D}^{1}_{5}=\left\{4=5^0*(5-1)\right\}$$ $$\mathfrak{D}^{2}_{5}=\left\{20=5^1*(5-1),24=5^1*(5-1)+5^0*(5-1)\right\}$$ $$\mathfrak{D}^{3}_{5}=\left\{100=5^2*(5-1),120=5^2*(5-1)+5*(5-1),124=5^2*(5-1)+5^1*(5-1)+5^0(5-1)\right\}$$
$$\mathfrak{D}^{4}_{5} = \{500,600,620,624\}$$

Here each element can be written as $\{500=5^3*(5-1),600=5^3*(5-1)+5^2*(5-1),620=5^3*(5-1)+5^2*(5-1)+5^1*(5-1),624=5^3*(5-1)+5^2*(5-1)+5^1*(5-1)+5^0*(5-1) \}.$\ Similarly $$\mathfrak{D}^{5}_{5} = \{2500,3000,3100,3120 \}.$$\ Here also each element can be written as $\{ 2500=5^5*(5-1),3000=5^5*(5-1)+5^2*(5-1),3100=5^5*(5-1)+5^4*(5-1)+5^3*(5-1),120=5^5*(5-1)+5^4*(5-1)+5^3*(5-1)+5^2*(5-1)+5^1*(5-1),3124=5^5*(5-1)+5^4*(5-1)+5^3*(5-1)+5^2*(5-1)+5^1*(5-1)+5^0*(5-1) \}$

Any pattern ?? .

Yes there is a pattern sir. For any $O^{n}-O^{m}$ for any odd number $O$ and any integers $n,m (m<n)$ can be expressed as :

$$O^{n}-O^{m}=\sum^{n-1}_{i=m} O^{i}*(O-1)$$

Now we can write the R.H.S of equation $$(p^{x}-p^{x-z}+3)=k(l+1)$$ as $$p^{x}-p^{x-z}= \sum^{x-1}_{x-z} p^i.(p-1)$$ After expanding the series and simplyifying we obtain $$p^{x-z}(p-1).\large\frac{p^x-1}{p-1}$$ $$p^{x-z}(p^z-1)$$

So we can equate to get $$p^{x-z}(p^z-1)=k(l+1)-3.$$

So we need to prove that both sides of the equation don't yield the same even number leading to a conrtradiciton. Hence I am trying that. I am here with a question again. Suppose we have an equation of this form $$p^{x-z}(p^{z}-1)=k(l+1)-3$$ where $p$ is an odd-prime, $z,y$ are integers and $y>0$ always ($x=2y+z,$ for some integer $x$ ) , $k$ is an odd number and $l$ is an even number.

So given such constraints and for any $p>3$ its well known that after substituting values of all variables its clear that both sides of the equation don't yield the same even number ( for any values ) . So how can we prove it ?.

I have been trying to prove it using congruences, but that didn't take me anywhere. So I wanted to ask it here.

Thank you.

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I don't understand what you mean. Is there a problem with p = 5, z = 1, y = 1, k = 3, l = 40, for example? Both sides yield 120 –  Cocopuffs Jun 14 '12 at 16:22
If I understand the problem correctly, the left side and the right side can hit the same even number. This is because the constraint $k(l+1)-3$ with $k$ odd and $l$ even says little about the right side. Example: Let $p=5$, $x=1$, $y=1$. The left side is $120$. Want $123=k(l+1)$, easy to arrange. –  André Nicolas Jun 14 '12 at 16:23
@Cocopuffs : So given that constraints can you comment now sir ? –  Iyengar Jun 14 '12 at 17:01
@ZevChonoles : Please don't close this question sir. I somehow clicked to close in hurry but later wanted to change this question. I have done it anyway. Thank you. –  Iyengar Jun 15 '12 at 18:03
@Iyengar There are ladies here too. I don't think it's appropriate call everybody "sir" or "Mr." –  user23211 Jun 22 '12 at 11:51

Edit: this only answers the old version of the problem. The question was changed so this answer is out of date.

The original problem was $$p^{x-z}(p^x-1)=\frac{n^2-1}{p^z}-3$$

Suppose we have a solution to the equation. The only hypotheses we need are that $p$ be any integer $\ge 2$, and $x>z>0$.

Then $$n^2=1+3p^z+p^x(p^x-1)$$ $$n^2=(p^x+1)^2-3(p^x-p^z)<(p^x+1)^2$$ $$n^2=(p^x-1)^2+p^x+3p^z>(p^x-1)^2$$ So we must have $$n^2=(p^x)^2$$ but this would imply $n^2=0 \pmod p$, which contradicts $n^2=1 \pmod p$ from the first equation above. So there cannot be any solution.

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Very clean answer sir, I didn't understand these things sir. 1) How do we obtain $\Delta^{\prime}=1+p^z(3+p^{2x-z}-p^{x-z})$ ?. I do know the general discriminant of $ax^2+bx+c$ which is $\Delta=b^2-4ac$ but how do we obtain that thing. What does reduced there mean ? . 2) How does $\Delta^'=(p^x)^2 \implies p^x-3p^z=1$ ? . Thanks a lot for your patience... –  Iyengar Jun 18 '12 at 6:04
(1) The reduced discriminant is $\Delta'=\Delta/4=(b/2)^2-ac$. This is useful when we know that $b$ is even. (2) Subtract the equality $\Delta'=(p^x)^2$ from the definition of $\Delta'$ and you get this equation. –  Generic Human Jun 18 '12 at 12:23
Sir how did you get $n^2=1+3p^z+p^x(p^x-1)$ in the first line. I don't understand that. –  Iyengar Jun 19 '12 at 16:56
Multiply the original equation by $p^z$ and add $1+3p^z$ to both sides. –  Generic Human Jun 19 '12 at 17:03
The original equation is $p^{x-z}(p^{z}-1)=\dfrac{ n^2-1}{p^z}-3$. But I made a mistake there.. so it now leads to a disturbance of the proof of Mr.Generic Human. –  Iyengar Jun 19 '12 at 18:38

First, if $x=z$ then $$(p^x+3)(p^x-1)+4 = p^{2x}+2p^x+1=(p^x+1)^2$$ which is always a square.

Second, the index is wrong in the sum $$p^{x}-p^{x-z}= \sum p^i(p-1)$$ it should be from $i=x-z$ to $x-1$, and after expanding the series and simplifying we get $$(p-1)p^{x-z} \sum^{z-1}_{i=0} p^i = p^{x-z}(p-1)\frac{p^z-1}{p-1} = p^{x-z}(p^z-1)$$ which we could see by factoring the left side.

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Yes sir you are right, Thanks a lot. –  Iyengar Jun 15 '12 at 5:08
Really a very good observation, Zander. Thanks a lot sir, for your modification... –  Iyengar Jun 15 '12 at 5:35
Since $p^z|n^2-1$, we could write $n^2-1=k(p^z)+m$ for some m,k where $m\lt p^z$, so we could plug this into the equation. Then we would get $p^{2y}(p^z-1)=k+{m\over p^2}-3$, since the left hand side is integer, we need the right hand side to be integer. However since $m\lt p^2$, the quantity $m\over p^2$ cannot be integer, so the equation does not hold.